Why is Hausdorff measure Borel regular?

Question

Definition of Hausdorff measure:

I already knew that all borel sets are measurable. So the problem is that given any subset $A$, how to find some borel set containing $A$ that has the same measure.

I have read some text, but they only say that we can replace the definition with open sets or closed sets and get the same definition (and this part I can understand), but then they claim that Hausdorff measure is borel regular as a corollary without explanation. Can any one give a detailed proof?

Thanks a lot.

Explanation of the fact we can replace the definition with open/closed subsets:

Answer 1

$\DeclareMathOperator{\diam}{diam}$First off, we can replace the $C_j$ with either open or closed sets for the reasons described in your second image. Since you seem to be a little confused on that point, let's look at that in a little more detail. First, suppose that we cover $A$ by a collection $C_j$. The diameter of the closure of a set is equal to the diameter of that set i.e. $\diam \overline{C}_j = \diam C_j$ for all $j$. But then $$\mathcal{H}_{\delta}^{m}(A) = C(m) \inf \sum_{j} (\diam C_j)^m = C(m) \inf \sum_{j} (\diam \overline{C}_j)^m, $$ where the infimum is taken over all $\delta$-coverings of $A$ (as above). Since size $\delta$ approximation of $\mathcal{H}^m(A)$ doesn't depend on whether or not the closures are closed, neither does the $m$-dimensional content.[1]

On the other hand, replacing the $C_j$ by open sets is slightly more delicate. However, it can be done: for any $\varepsilon > 0$, we can define a collection of sets of the form $$ C_{j,\varepsilon} := \left\{ x : d(x,C_j) < \frac{\varepsilon}{2^{j+1}} \right\} =: U_j, $$ where $d(x,C_j)$ denotes the distance from $x$ to $C_j$, i.e. $\inf_{y\in C_j} d(x,y)$. Notice that if we fatten up each of the $C_j$ into an open set, then (at worst) we are increasing the diameter by twice the amount of fattening-up. Hence $\diam U_j \le \diam C_j + \frac{\varepsilon}{2^j}$, so \begin{align} \mathcal{H}_{\delta}^{m}(A) &= C(m) \inf \sum_{j} (\diam C_j)^m \\ &\le C(m) \inf \sum_{j} (\diam U_j)^m \\ &\le C(m) \inf \sum_{j} \left[(\diam C_j)^m + \mathcal{O}\left( \frac{\varepsilon}{2^j} \right)\right] \\ &= \left[ C(m) \inf \sum_{j} (\diam C_j)^m\right] + \mathcal{O}(\varepsilon). \end{align} I'm sweeping a lot of details into that big-oh, so it would be a good idea to convince yourself that it is correct, and that I am not lying to you. The basic idea is that we can fatten up all of the sets in a cover by just a little bit in order to get an open cover. If we don't fatten things up too much, we end up with the same thing in the limit.

Alternatively, we can play the same game of fattening up a $\delta$-cover by a very small $\varepsilon$, then consider the $(\delta+\varepsilon)$-cover by open sets. Again, there are details that I am hiding, but you should be able to fill them in.

In short, we can replace the arbitrary $C_j$ in the original definition of the Hausdorff content with either open or closed $C_j$, and still get the same Hausdorff outer measure for any set.

This gets to the second part of your question: why does this imply that $\mathcal{H}^m$ is a regular Borel measure? It is usually a good idea to start with the definitions:

Definition: An outer measure $\mu$ is Borel if every Borel set $A$ is $\mu$-measurable, i.e. if $$ \mu(B) = \mu(A\cap B) + \mu(A\setminus B) $$ for any set $B$.

Showing that Hausdorff measure is Borel is nontrivial. The usual trick is to first show that Hausdorff measure is a metric outer measure, then invoke a theorem which states that all metric outer measures are Borel measures. I don't see how this particular property is a corollary of the fact that we can use either open or closed covers, but I'll sketch the proof here (I think that Folland's book on real analysis has a more complete proof, and one of Falconer's books almost certainly spells it out).

Definition: An outer measure $\mu^\ast$ is said to be a metric outer measure if $$ \mu^{\ast}(A\cup B) = \mu^{\ast}(A) + \mu^{\ast}(B) $$ whenever $\rho(A,B) > 0$, where $\rho(A,B)$ is the minumum distance between any two points in $A$ and $B$ (basically, we are requiring that $A$ and $B$ are contained in disjoint open sets; i.e. there is a fixed distance $\delta_0$ such that there are nonintersecting balls of radius $\delta_0$ centered at any two points in $A$ and $B$, respectively).

By construction $\mathcal{H}^m$ is an outer measure for any $m$ (we really only need to check subadditivity, which is not hard). On the other hand, if $A$ and $B$ are such that $\rho(A,B) = \delta_0$, then we can cover both $A$ and $B$ by countable collections of sets of radius $\min\{ \delta_0/3, \delta/2\}$ for any $\delta > 0$. Taking an infimum as $\delta \to 0$, we get the desired result.

Regularity, on the other hand, is a corollary of the fact that we can replace arbitrary covers with either open or closed covers. Recall:

Definition: $\mu$ is regular if for every set $A$ there exists an Borel set $B$ such that $A \subseteq B$ and $\mu(A) = \mu(B)$.

For each $n\in\mathbb{N}$, there exists some countable cover $\mathscr{U}_n = \{U_{n,j}\}$ of $A$ such that

1. Each set $U_{n,j} \in \mathscr{U}_n$ is open, and
2. $\sum_{j} (\diam U_{n,j})^m < \frac{1}{n}$.

Let $$ B := \bigcap_{n} \bigcup_{j} U_{n,j}. $$ By construction $B$ is Borel (it is a countable intersection of countable unions of open sets, therefore Borel—it probably even belongs to one of those fancy $G_{\sigma\delta}$ or $F_{\delta\sigma}$ classes of sets, but I can never remember the precise definitions of the sets in the hierarchy, so I won't embarrass myself by bringing up those kinds of sets. Oh... shoot.). Also note that $B$ has been built so that $$ \mu(A) = \mu(B), $$ which gives the regularity result. (Again, convince yourself that this is true.)

[1] Note that $C(m)$ is some constant that depends on $m$. Specifically, it is $\omega_m / 2^m$. I typically define the Hausdorff content without this constant, since it seems like a distraction to me, and can always be recovered later if need be.

Answer 2

For any subset $A \subset X$, take closed sets (e.g. closed balls) $E_{i,j}$ such that for every $i$ we have the properties:

1. $A \subset \cup_{j} E_{i,j}$ and
2. $d(E_{i,j}) \leq \frac{1}{i}$
3. $\sum_j \Big(\frac{d(E_{i,j})}{2}\Big)^m \leq \mathcal{H}^{m}_{\frac{1}{i}}(A) + \frac{1}{i}$

here $d(E)$ denotes the diameter of the set $E$.

Then, take $B=\cap_i \cup_j E_{i,j}$. $B$ is Borel because $E_{i,j}$ are closed, hence Borel. $A \subset B$, by property 1. Thus, $\mathcal{H}^{m}(A) \subset \mathcal{H}^{m}(B)$. Finally, by definition of $\mathcal{H}^m_{\delta}$ and by properties 1 and 2, and then property 3, this gives: $\mathcal{H}^m_{\frac{1}{i}}(B) \leq \mathcal{H}^{m}_{\frac{1}{i}}(A) + \frac{1}{i}$, and letting $i \to \infty$, we get $\mathcal{H}^{m}(B) \leq \mathcal{H}^{m}(A)$ as we wanted.